f(x)=ax2+bx+c(a>0,b∈R,c∈R) (1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求f(x)=ax2+bx+c(a>0,b∈R,c∈R)(1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求F(2)+F(-2)的值(2)
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![f(x)=ax2+bx+c(a>0,b∈R,c∈R) (1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求f(x)=ax2+bx+c(a>0,b∈R,c∈R)(1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求F(2)+F(-2)的值(2)](/uploads/image/z/5904330-42-0.jpg?t=f%28x%29%3Dax2%2Bbx%2Bc%28a%EF%BC%9E0%2Cb%E2%88%88R%2Cc%E2%88%88R%29+%EF%BC%881%EF%BC%89%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%98%AFf%28-1%29%3D0%2C%E4%B8%94c%3D1%2CF%28x%29%3Df%28x%29x%EF%BC%9E0%2C-f%28x%29x%EF%BC%9C0%2C%E6%B1%82f%28x%29%3Dax2%2Bbx%2Bc%28a%EF%BC%9E0%2Cb%E2%88%88R%2Cc%E2%88%88R%29%EF%BC%881%EF%BC%89%E8%8B%A5%E5%87%BD%E6%95%B0f%28x%29%E7%9A%84%E6%9C%80%E5%B0%8F%E5%80%BC%E6%98%AFf%28-1%29%3D0%2C%E4%B8%94c%3D1%2CF%28x%29%3Df%28x%29x%EF%BC%9E0%2C-f%28x%29x%EF%BC%9C0%2C%E6%B1%82F%282%29%2BF%EF%BC%88-2%EF%BC%89%E7%9A%84%E5%80%BC%EF%BC%882%EF%BC%89)
f(x)=ax2+bx+c(a>0,b∈R,c∈R) (1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求f(x)=ax2+bx+c(a>0,b∈R,c∈R)(1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求F(2)+F(-2)的值(2)
f(x)=ax2+bx+c(a>0,b∈R,c∈R) (1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求
f(x)=ax2+bx+c(a>0,b∈R,c∈R)
(1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求F(2)+F(-2)的值
(2)若a=1,c=0,且绝对值f(x)≤1在区间(0,1】恒成立,试求b取值范围
f(x)=ax2+bx+c(a>0,b∈R,c∈R) (1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求f(x)=ax2+bx+c(a>0,b∈R,c∈R)(1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求F(2)+F(-2)的值(2)
f(x)=x2+2x+1.这样第一问不可能做不出吧.
第二问【-2,0】分情况讨论,画图划出来就知道了
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证明二次函数f(x)=ax2+bx+c(a
证明二次函数f(x)=ax2+bx+c(a
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证明2次函数f(x)=ax2+bx+c(a>0)在区间[-b/2a,+∞)上是增函数
f(x)=ax2+bx+c(a>0,b∈R,c∈R) (1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求f(x)=ax2+bx+c(a>0,b∈R,c∈R)(1)若函数f(x)的最小值是f(-1)=0,且c=1,F(x)=f(x)x>0,-f(x)x<0,求F(2)+F(-2)的值(2)
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